Griffiths Introduction to Electrodynamics problem 3.21

Find the potential outside a charged metal sphere of charge $Q$ and radius $R$, placed in an otherwise uniform electric field $E_0\hat{z}$.

The Instructor's solution manual says that the electric potential has inverse $xy$-plane symmetry, so it can be solved as the same way as the uncharged sphere case, but I can't see why this symmetry holds.

In the uncharged sphere problem, inverse $xy$-symmetry holds since the upper/lower hemispheres have the same amount of charge, therefore having the same charge distribution. However, if the net charge is, say $Q>0$, positive charges will be concentrated at the upper hemisphere since there is fewer negative charges to resist to this movement. So I think the symmetry doesn't hold anymore. Is there anything wrong here?

//EDIT: inverse $xy$-symmetry=inverse symmetry of the $xy$-plane, $\rho$ at $(x,y,z)$ and $-\rho$ at $(x,y,-z)$.

  • $\begingroup$ One might indeed say that the charge density $\rho(z)$ is antisymmetric under reflection about the x,y plane. Note how the rotational symmetry of the problem eliminates the x,y dependence of $\rho$. $\endgroup$ Feb 27 at 0:35

1 Answer 1


The conducting sphere has an induced charge distribution with no net charge, and the external electric field is uniform and in the $\hat z$ direction. So the resultant field is such that it looks like the uniform case when far from the sphere, and when near it, turns in to meet the conducting surface in a way everywhere parallel to the surface normal. This problem has rotational symmetry about the z-axis. For example, it doesn't matter at which axis the x or y along which you approach the equator, the charge density looks the same in each case; this kind of symmetry holds for any slice parallel to the x,y plane; so indeed there is x,y symmetry (rotation about z-axis).

So, I looked up problem 3.8, and when Griffiths talks about using symmetry to set the whole x,y plane to zero, he is referring to the asymmetry of the charge density: $\rho(z)=-\rho(-z)$.

  • 1
    $\begingroup$ No, the conducting sphere is not uniformly charged. The charges in the surface of the metal is attracted by the electric field, forming an induced dipole. $\endgroup$
    – Leo
    Feb 26 at 0:06
  • $\begingroup$ @Leo Yes you are correct, the uniform electric field disperses the charge in the conductor; I think I was thinking about the electric field being uniform.. Second blunder I have made today; my allergies must be getting the best of me! I'll edit that off, everything else in the post holds though. $\endgroup$ Feb 26 at 2:29
  • $\begingroup$ Sorry, but I didn't mean rotation wrt z axis when I said $xy$-symmetry. I meant symmetry of the $xy$-plane, $(x,y,z)$ and $(x,y,-z)$. $\endgroup$
    – Leo
    Feb 27 at 0:17
  • $\begingroup$ @Leo Well, it is clear, as you know yourself, that there is no way the sphere will have a symmetric charge density when reflected across the x,y plane. If that is what the solution manual meant and not x,y independence, then it is certainly erroneous. $\endgroup$ Feb 27 at 0:22
  • $\begingroup$ When the net charge is 0, it has $xy$-symmetry. When there is charge density $\rho$ at $(x,y,z)$, there is $-\rho$ at $(x,y,-z)$. $\endgroup$
    – Leo
    Feb 27 at 0:27

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